# Fibonacci numbers extended

I am so excited and enjoyed the both the proofs of my previous question on Fibonacci series. I am so interested and fascinating person on fib series/functions. I use to do some rough work in my leisure time on the fib series. I encountered these new two following problems. As per my rough work and my calculator, I am very much correct to state the following questions, whereas mathematically, I am again failed to produce a proof of below. Can you help me please…

1. Let $f(x) = M(x) g(x)$ is a function with $M(x)$ is an f-even function and $g(x)$ is some continuous function. The product of f-even and continuous function is Fibonacci function if and only if continuous function should be a Fibonacci function. How to justify the above cited statement mathematically. Can we see such facts if we change f-even function to f-odd function of $M(x)$. If yes, what will happen?

2. Can we define the limit of the quotient of $f(x+1)/f(x)$, where $x$ tends to infinity? And the limit of this quotient is close to $1.6$, being the $f$ is Fibonacci sequence.

## Thank you so much for all members.

Fibonacci function I defined as per the request from mathematicians.

## Please look the definition of Fibonacci function at bellow my one of the comment and kindly answer my first question of this post.

Fibonacci function means, for all x in R, if f: R to R and satisfies f(x+2) = f(x+1) + f(x). Now, let me take f(x) = $M^x$ for some M > 0 As per my definition, f(x+2) = $M^x$ $M^2$ and f(x +1) = $M^x$ M and f(x) = $M^x$ Now, by Fibonacci function definition, we see; $M^2$ = M + 1 Which is same as (1 + square root 5)/2. Now our f(x) = $M^x$ = [(1 + square root of 5)/2]^x. I think, now one can answer my first question Thank you.

-
By Fibonacci function, do you mean generating function of Fibonacci sequence?? – Aang Jul 31 '12 at 6:09
I don't think you do mean the generating function of the Fibonacci sequence, which is a function $\hat f(x) = 1 + x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + \ldots$. I think you mean that $f(n)$ is the $n$th Fibonacci number, so for example $f(11)=89$. Otherwise your claim that the limit of $f(x+1)/f(x)$ is close to 1.6 makes no sense. – MJD Jul 31 '12 at 6:25
BMSA, I googled "fibonacci function f-even" and the only result relevant to this question ... was this question. If you're using made-up or little-known terminology, you need to provide clear definitions, unless you either want nobody to help you or you expect us to be mind-readers. If you want an explanation for the downvotes, there it is. I hope you can fix your question so that it makes sense so that it may be upvoted instead. – anon Jul 31 '12 at 6:31
You've edited your question, but you have left all the important stuff in the comments. Please edit into the question what you mean by Fibonacci function and by $f$-even function and anything else you have been asked about. – Gerry Myerson Aug 1 '12 at 7:28
You edited in the definition of Fibonacci function - good! You didn't edit in the definition of $f$-even - bad!! – Gerry Myerson Aug 1 '12 at 13:13

I don't know what "f-even" means, but it is well-known that $$\lim_{x\to\infty}{f(x+1)\over f(x)} = {1 + \sqrt 5 \over 2 } \approx 1.618\ldots$$

Wikipedia has a number of proofs of this and discussion of related matters. Here's a simple argument:

$$\begin{eqnarray} L & = & \lim_{x\to\infty}{f(x+1)\over f(x)} \\ & = & \lim_{x\to\infty}{f(x)+f(x-1)\over f(x)} \\ & = & \lim_{x\to\infty}1 + {f(x-1)\over f(x)} \\ & = & 1 + \lim_{x\to\infty}{f(x-1)\over f(x)} \\ & = & 1 + \lim_{x\to\infty}{f(x)\over f(x+1)} \\ & = & 1 + \lim_{x\to\infty}\left({f(x+1)\over f(x)}\right)^{-1} \\ & = & 1 + \left(\lim_{x\to\infty}{f(x+1)\over f(x)}\right)^{-1} \\ & = & 1 + L^{-1} \end{eqnarray}$$

Solving for $L$ in $L=1 + L^{-1}$ gives:

$$L={1 \pm \sqrt 5\over 2}$$

and we discard the superfluous negative solution $\frac12(1-\sqrt 5)$.

-
! f-even means f is some even function. Sir, I have seen wiki as per your reply. First of all prove that "limit exists" Then we can say 1.618... is the limit value. If possible, by using the "MONOTONE CONVERGENCE THEOREM". Thank you for quick reply. – BMSA Jul 31 '12 at 6:19
You are very welcome. I still don't understand "f-even". – MJD Jul 31 '12 at 6:20
! Wonderful solution you have given. In my school time, I learned the same kind of method in solving integration problems. Which is known as recursion method. Can we prove in some other method, other than your suggested method? I am looking some strong method. before that, how we can prove that EXISTENCE OF LIMIT by MONOTONE CONVERGENCE or some other method?But, I remember my school days. I love you to see your answer. Thank you so much. Hello...sir, can you think about my first question also... – BMSA Jul 31 '12 at 7:14
! the definition of even and odd is give above or see below: Let A(x) be a real valued function and for A(x) B(x) = 0 and B(x) is continuous then B(x) = 0. In this case map A(x) is said to be an f-even function for A(x+1) = A(x) and for an odd, A(x+1) = - A(x) for any real value of x. – BMSA Jul 31 '12 at 7:22

There are various ways to extend the notion of Fibonacci number. One of the more common ones is to let $$F(x)=\frac{\phi^x-\cos(\pi x)\phi^{-x}}{\sqrt{5}},$$ where $\phi=\frac{1+\sqrt{5}}{2}$. This gets over the difficulty that $\left(\frac{1-\sqrt{5}}{2}\right)^x$ makes no sense for most real values of $x$. Indeed one can in a similar way define $F(z)$ for complex $z$.

The required limit result is an almost immediate consequence of the definition.

-
Nicolas! your solution sounds good and little confusing. Can you explain little bit more clearly about F(z) please? – BMSA Jul 31 '12 at 7:23
@Nicolas! it seems quite interesting your thought. – BMSA Jul 31 '12 at 7:32
@BMSA. André is alluding to the fact that if you take $\phi =(1+\sqrt5)/2$ then $\phi^{-1}$ is a negative number and so the function $F(x)=(\phi^x-\phi^{-x})/\sqrt5$ will not be real-valued. His definition above fixes this problem. The real problem is that you haven't told us what you mean by "Fibonacci function". It'd be a very good idea if you did. – Rick Decker Jul 31 '12 at 13:55
@Rick Decker!Fibonacci function means, for all x in R, if f: R to R and satisfies f(x+2) = f(x+1) + f(x). Now, let me take f(x) = $M^x$ for some M > 0 As per my definition, f(x+2) = $M^x$ $M^2$ and f(x +1) = $M^x$ M and f(x) = $M^x$ Now, by Fibonacci function definition, we see; $M^2$ = M + 1 Which is same as (1 + square root 5)/2. Now our f(x) = $M^x$ = [(1 + square root of 5)/2]^x. I think, now you can answer my first question. – BMSA Aug 1 '12 at 4:36